Instrumental Chemical Analysis

Transcription

1 L15 Page1 Instrumental Chemical Analysis Nuclear Magnetic Resonance Dr. Ahmad Najjar Philadelphia University Faculty of Pharmacy Department of Pharmaceutical Sciences 1 st semester, 2017/2018

2 Nuclear Magnetic Resonance L15 Page2 Nuclear magnetic resonance ( NMR ) is a physical phenomenon based upon the magnetic properties of an atom's nucleus. All nuclei that contain odd numbers of nucleons and some that contain even numbers of nucleons have an intrinsic magnetic moment. The nuclei of many elemental isotopes have a characteristic spin ( I ), such as 1 H, 13 C, 15 N, 19 F, 29 Si and 31 P All nuclei with non-zero spins have magnetic moments (μ). A spinning charge generates a magnetic field, the resulting spin-magnet has a magnetic moment (μ) proportional to the spin ( I ). These nuclei absorb and re-emit electromagnetic radiation under the exposure of an external magnetic field.

3 Nuclear Magnetic Resonance L15 Page3 The energy (absorbed and re-emitted) is at a specific resonance frequency which depends on the strength of the magnetic field and the magnetic properties of the isotope of the atoms; in practical applications, the frequency is similar to VHF and UHF television broadcasts ( MHz).

4 Nuclear Magnetic Resonance L15 Page4 In the presence of an external magnetic field (B 0 ), two spin states exist, +1/2 and-1/2. The magnetic moment of the lower energy +1/2 state is aligned with the external field, but that of the higher energy -1/2 spin state is opposed to the external field.

5 Nuclear Magnetic Resonance NMR collects information concerning interactions between the nuclei of certain atoms present in the sample when they are submitted to a static magnetic field which has a very high and constant intensity and exposed to a second oscillating magnetic field. The second magnetic field, around times weaker than the first is produced by a source of electromagnetic radiation in the radiofrequency domain. L15 Page5 The difference between the number of protons in the low and high energy states is very small. This is because the energy difference between the two states is low relative to the thermal energy of the environment.

6 Nuclear Magnetic Resonance The wavelength of the radiation used in NMR is of low energy (radiofrequency region). The energy used in NMR absorption are expressed in Hertz, which is a unit of frequency. The stronger the magnetic field applied the greater difference in the populations of the higher (N 2 ) and lower (N 1 ) energy states (greater sensitivity) and the greater the radiation frequency in Hertz required to cause the spin of a nucleus to align against the field. for 1 H = rad s 1 T 1 h = J s 1 k = J. K 1 (for 1 H, R = if T =300K and B 0 = 5.3T) L15 Page6 The values for the strength of the applied magnetic field are in the range Gauss ( Tesla) NMR instruments are described in terms of the frequency at which they cause protons to resonate, thus a 600 mhz instrument is one which causes protons to resonate at frequency of 600 mhz.

7 Nuclear Magnetic Resonance L15 Page7 Instrumentation

8 Nuclear Magnetic Resonance L15 Page8 Instrumentation

9 L15 Page9 Instrumentation Source (RF transmitter): Nuclear Magnetic Resonance Radiofrequency (RF) transmitters generate frequencies of a few MHz to almost 1 GHz, which irradiate the sample molecules. If the energy difference between the relevant spin states is matched by the RF pulse, the nuclei will move to the higher spin state and be in resonance with the magnetic field. In older instruments, either the frequency sweep or field sweep methods are used. In the frequency sweep method, the magnetic field is held constant while the frequency is continuously varied until the resonant condition is met for each nuclide. In the field sweep method, the frequency is held constant while the magnetic field is varied. More recently, pulsed NMR has taken over in which RF transmitters provide pulses of radiation to simultaneously excite all of the one type of nuclide, e.g. carbon-13 in the molecule.

10 Nuclear Magnetic Resonance L15 Page10 Instrumentation Sample: The sample is placed in the probe which is placed in the centre of the bore. At the top of the probe are coils surrounded by electronics and tuning components. There are also now flow probes, solid state probes and cryogenic (cooled) probes. The cryogenic probes cool the electronics to reduce background noise, which can improve the S/N ratio by 3 4 times.

11 Nuclear Magnetic Resonance L15 Page11 Instrumentation Discriminator: The magnetic field is the discriminator. The vast majority of magnets used today are super-conducting cryomagnets, which allow higher field strengths than older permanent magnets or electromagnets. These cryomagnets contain a large solenoid immersed in liquid helium at a temperature of 4.2 K or below. The coil resides inside a liquid helium Dewar flask at the centre of the magnet through which the bore passes. The bore is at room temperature. To reduce boil-off of the helium, the Dewar flask is surrounded by a second Dewar flask containing liquid nitrogen. Some magnets include a second superconducting coil outside the main solenoid to cancel out the effects of the field in the surrounding room. Supercooled magnets can obtain field strengths of up to 21 T (900 MHz).

12 Nuclear Magnetic Resonance L15 Page12 Instrumentation Detector (RF receiver): When the resonant condition is met, the NMR signal is collected at the RF receivers. NMR signals are generally weak and need to be amplified and processed prior to further analysis. Using the pulsed mode, the free induction decay (FID) spectrum in the time domain is recorded and while it contains all the information on frequencies, splitting and integrals, it must be converted into the frequency domain by Fourier Transformation (FT). This FT step enhances the S/N ratio of the signal

13 Nuclear Magnetic Resonance L15 Page13 Instrumentation Output At the computer, the huge amount of information is processed and spectral searching and matching can be carried out. NMR spectra can be very complex, especially two dimensional (2-D) experiments, and may require detailed data analysis and interpretation. Libraries can be very useful for assigning structure and identifying compounds.

14 Nuclear Magnetic Resonance L15 Page14 1 H-nmr spectrum Information from 1 H-nmr spectrum: 1. Number of signals: How many different types of hydrogens in the molecule. 2. Position of signals (chemical shift): What types of hydrogens. 3. Relative areas under signals (integration): How many hydrogens of each type. 4. Splitting pattern (multiplicity): How many neighboring hydrogens.

15 Nuclear Magnetic Resonance L15 Page15 1 H-nmr spectrum

16 Nuclear Magnetic Resonance L15 Page16 1 H-nmr spectrum: Number of signals Magnetically equivalent hydrogens resonate at the same applied field. Magnetically equivalent hydrogens are also chemically equivalent. # of signals?

17 L15 Page17 Nuclear Magnetic Resonance 1 H-nmr spectrum: Position of signals (Chemical shift) Nuclei are surrounded by electrons. The strong applied magnetic field (B o ) induces the electrons to circulate around the nucleus. The induced circulation of electrons sets up a secondary (induced) magnetic field (B i ) that opposes the applied field (B o ) at the nucleus. We say that nuclei are shielded from the full applied magnetic field by the surrounding electrons because the secondary field diminishes the field at the nuclei. The electron density surrounding a given nucleus depends on the electronegativity of the attached atoms. The more electronegative the attached atoms, the less the electron density around the nucleus in question.

18 L15 Page18 Nuclear Magnetic Resonance 1 H-nmr spectrum: Position of signals (Chemical shift) We say that that nucleus is less shielded, or is deshielded by the electronegative atoms. Deshielding effects are generally additive. That is, two highly electronegative atoms (2 Cl atoms, for example) would cause more deshielding than only 1 Cl atom. We define the relative position of absorption in the NMR spectrum the chemical shift. It is a unitless number (actually a ratio, in which the units cancel), but we assign units of ppm or (Greek letter delta) units. For 1 H, the usual scale of NMR spectra is 0 to 10 (or 12) ppm (or ). The zero point is defined as the position of absorption of a standard, tetramethylsilane (TMS): This standard has only one type of C and only one type of H.

19 Nuclear Magnetic Resonance L15 Page19 1 H-nmr spectrum: Position of signals (Chemical shift) C C O C H O C OH O C H Aromatic H C C H O C C CH3 X C H H Ar CH 3 C C H CH 2 CH CH 3 TMS downfield H Chemical shift ( ) 4 3 upfield 2 1 0

20 Nuclear Magnetic Resonance L15 Page20 1 H-nmr spectrum: Position of signals (Chemical shift)

21 Nuclear Magnetic Resonance L15 Page21 1 H-nmr spectrum: Relative areas under signals Area α Number of H atoms, area ratios between different types of H atoms equal the ratios between the numbers of these atoms Example: C 5 H 12 O isomers H 3 C a CH 3 CH 3 b OH c Signals Value Rel. Area of Signal a ~ 1 9 b > 2 2 Low Resolution NMR c ~ b 2 c 1 a 9 tms Low Resolution NMR a 9 H 3 C a CH 3 O CH 3 CH 3 b Signals Value Rel. Area of Signal a ~ 1 9 b >>1 3 b tms

22 Nuclear Magnetic Resonance L15 Page22 1 H-nmr spectrum: Relative areas under signals

23 Nuclear Magnetic Resonance L15 Page23 1 H-nmr spectrum: Relative areas under signals

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25 Nuclear Magnetic Resonance L15 Page25 1 H-nmr spectrum: Spin-Spin Splitting Spin-spin splitting occurs only between nonequivalent protons on the same carbon or adjacent carbons.

26 Nuclear Magnetic Resonance L15 Page26 1 H-nmr spectrum: Spin-Spin Splitting Why are There Three Lines for H B? H B feels the splitting of both H A and H A. So, let s imagine starting with H B as a single line, then let s turn on the coupling from H A and H A one at a time: H B If uncoupled, H B would appear as a singlet where the dashed line indicates the chemical shift of the singlet. Now, let's "turn on" H B - H A coupling. This splits the single line into two lines H A H B Now, let's "turn on" H B - H A' coupling. This splits each of the two new lines into two lines, but notice how the two lines in the middle overlap. Overall, we then have three lines. H A' C C Because the two lines in the middle overlap, that line is twice as big as the lines on the outside. More neighboring protons leads to more lines as shown on the next slide.

27 Nuclear Magnetic Resonance L15 Page27 1 H-nmr spectrum: Spin-Spin Splitting Coupling constant (J): The separation on an NMR spectrum (in hertz) between adjacent peaks in a multiplet

28 Nuclear Magnetic Resonance L15 Page28 1 H-nmr spectrum: Spin-Spin Splitting

29 Nuclear Magnetic Resonance L15 Page29 1 H-nmr spectrum: Spin-Spin Splitting More complex spin-spin coupling: non equivalent protons will couple independently. Summary of 1H-1H spin-spin coupling Chemically equivalent protons do not exhibit spin-spin coupling to each other. The resonance of a proton that has n equivalent protons on the adjacent carbon is split into n+1 peaks (multiplicity) with a coupling constant J. Protons that are coupled to each other have the same coupling constant Non-equivalent protons will split a common proton independently. complex coupling. Spin-spin coupling is normally observed between nuclei that are one, two and three bonds away. Fourbond coupling can be observed in certain situations but is not common.

30 Nuclear Magnetic Resonance L15 Page30 1 H-nmr spectrum: Chemical Shift

31 L15 Page31 Type of Chemical Hydrogen Shift ( ) ( CH 3 ) 4 Si 0 (by definition) RCH RCH 2 R R 3 CH R 2 C= CRCHR RC CH ArCH ArCH 2 R ROH RCH 2 OH RCH 2 OR R 2 NH O RCCH O RCCH 2 R Type of Hydrogen O RCOCH 3 O RCOCH 2 R RCH 2 I RCH 2 Br RCH 2 Cl RCH 2 F ArOH R 2 C= CH 2 R 2 C= CHR ArH O RCH O RCOH Chemical Shift ( )

32 L15 Page32 Factors affected Chemical shift : (1) electronegativity of nearby atoms, (2) hybridization of adjacent atoms, and (3) diamagnetic effects Electronegativity CH 3 -X CH 3 F CH 3 OH CH 3 Cl CH 3 Br CH 3 I (CH 3 ) 4 C (CH 3 ) 4 Si Electron egativity of X Chemical Shift ( )

33 L15 Page33 Hybridization of adjacent atoms Type of Hydrogen (R = alkyl) RCH 3, R 2 CH 2, R 3 CH R 2 C=C(R)CHR 2 Name of Hydrogen Alkyl Allylic Chemical Shift ( ) RC CH Acetylen ic R 2 C=CHR, R 2 C=CH 2 RCHO Vin ylic Ald ehydic

34 Interpreting NMR Spectra L15 Page34 Alkanes 1 H-NMR signals appear in the range of Alkenes 1 H-NMR signals appear in the range H-NMR coupling constants are generally larger for trans-vinylic hydrogens (J= Hz) compared with cis-vinylic hydrogens (J= 5-10 Hz). Alcohols 1 H-NMR O-H chemical shift often appears in the range , but may be as low as H-NMR chemical shifts of hydrogens on the carbon bearing the -OH group are deshielded by the electron-withdrawing inductive effect of the oxygen and appear in the range Ethers A distinctive feature in the 1 H-NMR spectra of ethers is the chemical shift, , of hydrogens on the carbons bonded to the ether oxygen.

35 L15 Page35 Aldehydes and ketones 1 H-NMR: aldehyde hydrogens appear at H-NMR: a-hydrogens of ketones appear at Amines 1 H-NMR: amine hydrogens appear at depending on conditions. Carboxylic acids 1 H-NMR: carboxyl hydrogens appear at ppm, higher than most other types of hydrogens.

36 L15 Page36 a c b a c c a b c b a b a b c c a b

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41 L15 Page41 13 C NMR Spectroscopy Nuclear Magnetic Resonance 13 C is a much less sensitive nuclei than 1 H for NMR spectroscopy Chemical shifts give an idea of the chemical and electronic environment of the 13 C nuclei due to shielding and deshielding effects range: ppm from TMS 13 C NMR spectra will give a map of the carbon framework. The number of resonances equales the number of non-equivalent carbons. 1 H- 13 C spin-spin coupling: spin-spin coupling tells how many protons are attached to the 13 C nuclei. (i.e., primary, secondary tertiary or quaternary carbon) 13 C spectra are usually collected with the 1 H- 13 C coupling turned off (broad band decoupled). In this mode all 13 C resonances appear as singlets. Chemical Shift Range of 13 C

42 Nuclear Magnetic Resonance L15 Page42 13 C NMR Spectroscopy

43 L15 Page43 13 C NMR Spectroscopy Nuclear Magnetic Resonance DEPT spectra (Distortionless Enhancement by Polarization Transfer) is a modern 13 C NMR spectra that allows you to determine the number of attached hydrogens. Run: broad-band decoupled spectra DEPT-90: only CH s show up DEPT-135: CH s and CH 3 s give positive resonances CH 2 s give negative resonances

44 Practice L15 Page44

45 How 1 H NMR spectrum for p-xylene will appear? L15 Page45 p-xylene H 3 C CH 3 How many H type are their? This will inform us the number of signals in the spectra How many H atoms in each type? This will inform us the area of each signal What are the functional groups near each H type? This will inform us where the signal will appear (chemical shift ) How many H atoms are attached to the carbon atom s nearby each H type? This will inform us how the splitting pattern will appear for each signal

46 How 1 H NMR spectra for p-xylene will appear? L15 Page46 p-xylene H 3 C CH 3 a b a a 6H at about 2.2 ppm singlet b 4H at about 7.0 ppm singlet

47 How 1 L15 Page47 H NMR spectra for tert-butyl bromide will appear? CH 3 tert-butyl bromide H 3 C C CH 3 Br a singlet 9H at about 1.8ppm

48 How 1 L15 Page48 H NMR spectra for ethyl bromide will appear? a b ethyl bromide CH 3 CH 2 -Br a triplet 3H b quartet 2H

49 Draw the chemical structure for the compound L15 Page49 Number of signals in the spectra Area of each signal Chemical shift of the signal Splitting pattern for each signal will inform us the number of H type in the molecule will inform us the number of H atoms in each type will inform us about closeness of H type to the functional group will inform us how many H atoms surrounding each type

50 Draw the chemical structure for the compound C 3 H 7 Br a triplet 3H b complex 2H c triplet 2H L15 Page50 c b a

51 Draw the chemical structure for the compound C 3 H 7 Br o C 3 H 7 Br has IHD = 0 no unsaturation (no double bonds nor cycles), so it is chain molecule (straight or branched). o From the given data, there are three signals (a, b and c) so there are three types of H atoms, each type will be at a unique carbon atom. o It is obvious that there is no symmetry in the molecule, the formula has 7H atoms and they are distributed on three separate signals (3+2+2). We have CH 3 -, -CH 2 - and -CH 2 - a b c o c signal has the higher value (more deshielded atoms), this mean that this H type of atoms are the nearest to the Br atom. We could use tables to check for this, so c signal should be like this -CH 2 -Br c o The splitting pattern shows that a signal is triplet, this mean that it has only 2H atoms surrounding it, so this side of the molecule will be CH 3 -CH 2 - because it has only one side to attach with. a o Now, the final touch, the like between the tow ends, it is the b hydrogens. This will form CH 3 -CH 2 -CH 2 -Br. This justify the splitting patterns for b and c. a b c L15 Page51

52 L15 Page52 Draw the chemical structure for the compound C 3 H 7 Br 1-bromopropane a b c CH 3 CH 2 CH 2 -Br a triplet 3H b complex 2H c triplet 3H

53 Try the same methodology for C 3 H 7 Cl spectra below a doublet 6H b septet 1H L15 Page53 b a Answer: isopropyl chloride

54 L15 Page54 Draw the chemical structure for the compound C 4 H 9 Br 2-bromobutane b d c a CH 3 CHCH 2 CH 3 Br a triplet 3H b doublet 3H c complex 2H d sextet 1H

55 L15 Page55 Draw the chemical structure for the compound C 8 H 9 Cl o-methylbenzyl chloride a CH 3 b CH 2 Cl c a singlet 3H b singlet 2H c ~ singlet 4H

56 L15 Page56 Draw the chemical structure for the compound C 2 H 6 O ethanol a c b CH 3 CH 2 -OH a triplet 3H b singlet 1H c quartet 2H

57 Draw the chemical structure for the compound C 8 H 10 L15 Page57 ethylbenzene c b a CH 2 CH 3 a triplet 3H b quartet 2H c ~singlet 5H

58 Draw the chemical structure for the compound C 10 H 14 p-diethylbenzene a b c b a CH 3 CH 2 CH 2 CH 3 L15 Page58 a triplet 6H b quartet 4H c singlet 4H

59 Draw the chemical structure for the compound C 10 H 14 L15 Page59 o-diethylbenzene c b a CH 2 CH 3 a triplet 6H b quartet 4H c singlet 4H CH 2 CH 3 b a

60 Draw the chemical structure for the compound C 10 H 14 L15 Page60 m-diethylbenzene

61 L15 Page61 Draw the chemical structure for the compound C 5 H 11 Br 2-bromo-2-methylbutane b CH 3 b CH 3 CCH 2 CH 3 a Br c a triplet 3H b singlet 6H c quartet 2H b & c overlap

62 L15 Page62 Draw the chemical structure for the compound C 6 H 14 O di-n-propylether a b c c b a CH 3 CH 2 CH 2 -O-CH 2 CH 2 CH 3 a triplet 6H b complex 4H c triplet 4H

63 L15 Page63 Draw the chemical structure for the compound C 3 H 8 O 1-propanol a b d c CH 3 CH 2 CH 2 -OH a triplet 3H b complex 2H c singlet 1H d triplet 2H

64 C 11 H 16 L15 Page64 a 9H = 3CH 3, no neighbors c 5H = monosubstituted benzene b 2H, no neighbors c b a 9H CH 3 CH 2 C CH 3 CH 3 5H neopentylbenzene 2H

65 C 4 H 8 Br 2 L15 Page65 a = 6H, two CH 3 with no neighbors (CH 3 ) 2 C b = CH 2, no neighbors & shifted downfield due to Br 6H H 3 C CH 3 C CH 2 Br Br 2H

66 C 7 H 8 O L15 Page66 H 2 C OH 5H c = monosubst. benzene b = CH 2 c = OH 2H 1H

67 C 4 H 9 Br a doublet 1.04 ppm b complex 1.95 ppm c doublet 3.33 ppm 6H 1H 2H L15 Page67 a = two equivalent CH 3 s with one neighboring H (b?) c = CH 2 with one neighbor H (also b) a CH 3 a 6H doublet CH 3 CHCH 2 Br b 1H complex a b c c 2H doublet

68 L15 Page68 C 10 H 13 Cl a singlet 1.57 ppm 6H b singlet 3.07 ppm 2H c singlet 7.27 ppm 5H a = two-equilalent CH 3 s with no neighbors c = monosubstituted benzene ring b = CH 2 c b CH 3 CH 2 C CH 3 Cl a a singlet 6H b singlet 2H c singlet 5H